Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As ch

Question

Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: a 2.50-kg stone thrown upward from the ground at 15.0 m/s returns to the ground in 9.00 s ; the circumference of Mongo at the equator is 2.00×10^5 KM; and there is no appreciable atmosphere on Mongo.

1-The starship commander, Captain Confusion, asks for the following information: what is the mass of Mongo?

2-If the Aimless Wanderer goes into a circular orbit 2.00×10^4 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?

Explanation / Answer

1. If the stone takes 9s to get back to the ground, then let's use kinematics to find its acceleration:

vo = 15m/s, v = -15m/s, t = 9s
v = vo + at
-15m/s = 15m/s + at
a = -3.333m/s or 3.333m/s downward

Now, let's use newton's second and the universal law of gravitation to find the mass of the planet

F = ma

GmM/r^2 = m*a

m's cancel

GM/r^2 = a

convert your radius from km to m and solve:

M = a*r^2/G = (3.333m/s^2)(2*10^8m)^2/(6.67*10^-11) = 1.999*10^27 kg

2. Orbiting 2*10^4km above the surface would be (2*10^4km + 2*10^5km) = 2.2*10^5km from the center. So:

F = ma

GmM/r^2 = mv^2/r

m's and one r cancel

GM/r = v^2

again remember to convert km to m

v^2 = (6.67*10^-11)(1.999*10^27kg)/(2.2*10^8)

v = 24,618 m/s

For circular motion:

v = (circumference)/(period) = 2r/T

T = 2r/v = 2(2.2*10^8m)/(24,618m/s) = 56,149s

or 15.6 hours

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